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Given an array of integers and a number k, find knon-overlapping subarrays which have the largest sum.
The number in each subarray should be contiguous.
Return the largest sum.
Given [-1,4,-2,3,-2,3], k=2, return 8
Note
The subarray should contain at least one number
Analysis:
DP. d[i][j] means the maximum sum we can get by selecting j subarrays from the first i elements.
d[i][j] = max{d[p][j-1]+maxSubArray(p+1,i)}
we iterate p from i-1 to j-1, so we can record the max subarray we get at current p, this value can be used to calculate the max subarray from p-1 to i when p becomes p-1.
1 class Solution { 2 public: 3 /** 4 * @param nums: A list of integers 5 * @param k: An integer denote to find k non-overlapping subarrays 6 * @return: An integer denote the sum of max k non-overlapping subarrays 7 */ 8 int maxSubArray(vector<int> nums, int k) { 9 // write your code here 10 int n = nums.size(); 11 vector<vector<int>> dp(n + 1, vector<int>(k + 1, INT_MIN)); 12 for (int i = 0; i <= n; ++i) 13 dp[i][0] = 0; 14 for (int i = 1; i <= n; ++i) { 15 for (int j = 1; j <= min(i, k); ++j) { 16 int tmp = 0; 17 for (int t = i - 1; t >= j - 1; --t) { 18 tmp = max(tmp + nums[t], nums[t]); 19 dp[i][j] = max(dp[i][j], tmp + dp[t][j-1]); 20 } 21 } 22 } 23 return dp[nums.size()][k]; 24 } 25 };
重复使用dp数组，可以将空间复杂度降到O(n)。为了避免冲突，在枚举求解数组长度n时到倒着求，这样可以保证上一次迭代的结果不被覆盖掉。
1 class Solution { 2 public: 3 /** 4 * @param nums: A list of integers 5 * @param k: An integer denote to find k non-overlapping subarrays 6 * @return: An integer denote the sum of max k non-overlapping subarrays 7 */ 8 int maxSubArray(vector<int> nums, int k) { 9 // write your code here 10 int n = nums.size(); 11 vector<int> dp(n + 1, 0); 12 for (int j = 1; j <= k; ++j) { 13 for (int i = n; i >= j; --i) { 14 dp[i] = INT_MIN; 15 int tmp = 0; 16 for (int t = i - 1; t >= j - 1; --t) { 17 tmp = max(tmp + nums[t], nums[t]); 18 dp[i] = max(dp[i], tmp + dp[t]); 19 } 20 } 21 } 22 return dp[n]; 23 } 24 };
[LintCode] Maximum Subarray III
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原文地址：http://www.cnblogs.com/easonliu/p/4566954.html
